Current Electricity question for JEE Main (JEE Main 2023 (11 Apr, Shift 2)), with solution

Answer: (C) $\dfrac13\ \text{A}$

$R_6$ is directly across the cell and does not affect the rest.

Between $C$ and $A$: $R_3=6\ \Omega$ in parallel with $R_2+R_1=6\ \Omega$, i.e. $3\ \Omega$. Then $D\to C\to A$ is $3+3=6\ \Omega$, in parallel with $R_7=3\ \Omega$: $2\ \Omega$.

With $R_5$: $2+2=4\ \Omega$, so the current from $E$ is $\dfrac84=2\ \text{A}$ and $V_D-V_A=2\times2=4\ \text{V}$.

Current $D\to C\to A$: $\dfrac46=\dfrac23\ \text{A}$, which splits equally at $C$: $I_{R_2}=\dfrac13\ \text{A}$.

Solution by Sreeraj P, M.Sc Physics